Computer Graphics BCA 303

Chanderprabhu Jain College of Higher Studies & School of Law
Plot No. OCF, Sector A-8, Narela, New Delhi – 110040
(Affiliated to Guru Gobind Singh Indraprastha University and Approved by Govt of NCT of Delhi & Bar Council of India)
PAPER NAME: COMPUTER GRAPHICS
PAPER CODE: BCA 303
CLASS: BCA Vth Semester

UNIT-I Graphics Display Devices
• Frame Buffer – a region of memory sufficiently large to hold all of the
pixel values for the display
Block
Diagram
of
Computer
with
Raster
Display
(Affiliated to Guru Gobind Singh Indraprastha University and Approved by Govt of NCT of Delhi & Bar Council ofIndia)2

Graphics Display Devices - cont
• How each pixel value in the frame buffer
is sent to the right place on the display
surface

Graphics Devices – cont
• Each pixel has a 2D address (x,y)
– For each address (x,y) there is a specific memory
location that holds the value of the pixel (I.e.
mem[136][252])
– The scan controller sends the logical address (136,
252) to the frame buffer, which emits the value
mem[136][252]
– The value mem[136][252] is converted to a
corresponding intensity or color in the conversion
circuit, and that intensity or color is sent to the
proper physical position, (136, 252), on the display
surfaceChanderprabhu Jain College of Higher Studies & School of Law

mem[136][252])
mem[136][252]

Scan Converting Lines
• Line Drawing
– Draw a line on a raster screen between 2
points
– What’s wrong with the statement of the problem?
• It does not say anything about which pts are
allowed as end pts
• It does not give aclear meaning to“draw”
• It does not saywhat constitutes a“line”in the raster
world
• It does not say how to measure the success of the
proposed algorithm

mem[136][252])
mem[136][252]
(Affiliated to Guru Gobind Singh Indraprastha University and Approved by Govt of NCT of Delhi & Bar Council ofIndia1)0

Scan Converting Lines - cont
• Problem Statement
– Given 2 points P and Q in the plane, both
with integer coordinates, determine which
pixels on a raster screen should be “on” in
order to make a picture of a unit-width line
segment starting at point P and ending at
point Q

Finding the next pixel
• Special Case:
– Horizontal Line:
• Draw pixel P and increment the x coordinate value by one
to get the next pixel.
– Vertical Line:
• Draw the pixel P and increment the y coordinate value by one
to
get the next pixel
– Diagonal Line:
• Draw the pixel P and increment both the x and y coordinate
values by one to get the next pixel
– What should we use in the general case?

Vertical Distance
• Why can we use the vertical
distance as a measure of which
point is closer?
– Because vertical distance is
proportional to the actual distance
– How do we show t
– Congruent Triangl
his
?
es

Vertical Distance – cont
• By similar triangles we can see that the
true distances to the line (in blue) are
directly proportional to the vertical
distances to the line (in black) for each
point.
• Therefore the point with the smaller
vertical distance to the line is the
closest to the line

Strategy 1 – IncrementalAlgorithm
• The Basic Algorithm
– Find the equation of the line that
connects the 2 points P and Q
– Starting with the leftmost point P,
increment where A
=
xi
by 1 to calculate
slope, and B = y
intercept
– Intensify the pixel at
yi  Axi  B– This computation selects the closest pixel,
the pixel whose distance to the “true” line is
smallest

Strategy 1 – IncrementalAlgorithm
• The Incremental Algorithm
– Each iteration requires a floating-
point multiplication therefore,
modify
Yi  1  Axi  1  B  Axi  x B  yi
– If , then– Thusx,a1unit chayni 
g1
einyi
xchAanges y by
slope A, which is the slope of the line
– At each step, we make incremental
calculations based on the preceding step
to find the next y value
 Ax

Strategy 1 – Incremental Algo

Example Code

Problem with the IncrementalAlgorithm
• Rounding integers takes time
• Real variables have limited precision,
summing an inexact slope (A) repetitively
introduces a cumulative error buildup
• Variables y and A must be a real or
fractional binary because the slope is a
fraction
– Special case needed for vertical lines

Strategy 2 – Midpoint LineAlgorithm
• Assumethat the line’s slope is shallow and positive ( 0<
slope < 1); other slopes can be handled by suitable
reflections about the principle axes
• Call the lower left
endpoint
and the
upperright
endpoint
x0, y0
 • Assume that we hxa1
v,eyju1
st selected the pixel P at
• Next, we must choose between the pixel to the
right (pixel E), or one right and one up (pixel
NE)
• Let Q be the intersection point of the line being
scan- converted with the grid line
xp, yp 
Chanderprabhu Jain College oxf HighxeprStu1dies & School of Law

Strategy 2 – Midpoint Line Algorithm
1/1/2000

Strategy 2 – Midpoint LineAlgorithm
• The line passes between E and NE
• The point that is closer to the intersection point Q
must be chosen
• Observe on which side of the line the midpoint M lies:
– E is closer to the line if the midpoint lies above the line (I.e.
the line crosses the bottom half)
– NE is closer to the line if the midpoint lies below the line, I.e., the
line
crosses the top half
• The error, thevertical distance between the chosen pixel
and the actual line is always <= ½
• The algorithm chooses NE as the next pixel for the line
shown
• Now, find a way to calculate on which side of the line the
midpoint lies

The Line
• The line equation as a function f(x):
– f(x) = A*x + B = dy/dx * x + B
• Line equation as an implicit function:
– F(x,y) = a * x + b * y + c = 0 for coefficients a, b, c where a, b !=
0; from above, y *dx = dy*x + B*dx, so a = dy, b = -dx,
c=B *dx, a>0 for y(0) < y(1)
• Properties (proof by the case analysis):
–
–
when any point M is on the line
when any point M is above the
line– F(xm, ym)  0 when any point M is below the line
– OurFde(xcimsi,oynmw)illb0ebased on the value of the function at the
midpoint
M aFt (xm, ym) 0
2
1x 1, y 
(Affiliated to Guru Gobind Singh Indraprpastha Universitpyand Approved by Govt of NCT of Delhi & Bar Council of India2)3

Decision Variable
• Decision Variable d:
to see
where
• How to update d:
– On the basis of picking E or NE, figure out the
location of the M for that pixel, and the
corresponding value of d for
the next grid line
– We only need the sign of F (xp 1, yp 
1
2)the line lies, and then pick the nearest
pixel
– D  F (xp 1, yp 1
2)
• If d > 0 choose pixel NE
• If d < 0 choose pixel E
• If d = 0 choose either one consistently

Example Code

Rotation About the Origin
To rotate a line or polygon, we must rotate
each of its vertices.
To rotate point (x1,y1) to point (x2,y2) we
observe:
From the illustration we know that:
sin (A + B) = y2/r cos (A + B) = x2/r
sin A= y1/r cos A =x1/r
x-axis
(x1,y1)
(x2,y2)
A
B
r
(0,0)
y-axis
UNIT-II 2-D TRANSFORMATIONS
(Affiliated to Guru Gobind Singh Indraprastha University and Approved by Govt of NCT of Delhi & Bar Council ofIndia)

Rotation About the Origin
From the double angle formulas: sin (A + B) =
sinAcosB + cosAsinB
cos (A + B)= cosAcosB -
sinAsinB
y2/r = (y1/r)cosB +Substituting:
(x1/r)sinB
Therefore: y.2= yP11cosB+ x1sinB
We have
y1sinB
P2 = R
(y2) (sinB cosB) (y1)
(x2) (cosB -sinB) (x1)x2 = x1cosB -
=

Translations
y( t )1( y )
P2 = (x1 + tx)
(y1 + ty)
Moving an object is called a translation. We translate a point by
adding to the x and y coordinates, respectively, the amount the
point should be shifted in the x and y directions. We translate an
object by translating each vertex in the object.
P2 = P1 + T
Ty
P1 = ( x1 ) T = ( tx )
Tx

Scaling
Changing the size of an object is called a scale. We scale an object by
scaling the x and y coordinates of each vertex in the object.
P2 = .SP1
S = (sx
(0 sy)
0) P1 = ( x1 )
)( y1
P2 =(sxx1)
(syy1)

Homogeneous Coordinates
cos ø -sin ø 0
Rø = sin ø cos ø 0
0 0 1
Sx 0 0
S = 0 Sy 0
0 0 1
1 0 T
x
T = 0 1 T
y
0 0 1
Although the formulas we have shown are usually the most efficient way to
implement programs to do scales, rotations and translations, it is easier to use
matrix transformations to represent and manipulate them.
In order to represent a translation as a matrix operation we use 3 x 3 matrices
and pad our points to become 1 x 3 matrices.
(x)
Point P =(y)
(1)

Composite Transformations - Scaling
Given our three basic transformations we can create other
transformations.
Scaling with a fixed point
A problem with the scale transformation is that it also moves the
object being scaled.
Scale a line between (2, 1) (4,1) to twice its length.
(2 0 0)
(0 1 0)
(0 0 1)
(2 0 0)
(0 1 0)
(0 0 1)
=
=
Before
After
(2) (4)
(1) (1)
(1) (1)
(4) (8)
(1) (1)
(1) (1)
Chanderprabhu Jain Colle0ge 1of H2 ig3her4Stu5die6s &7 Sc8ho9olo10fLaw

Composite Transforms - Scaling (cont.)
If we scale a line between (0, 1) (2,1) to twice its length, the left-
hand endpoint does not move.
=(2 0 0)
(0 1 0)
(0 0 1
)
(0 0 1) (1)
(2 0 0) (1) =
(0 1 0) (2)
(0,0) is known as a fixed point for the basic scaling transformation.
We can used composite transformations to create a scale
transformation with different fixed points.
0 1 2 3 4 5 6 7 8 9 10
Before
After
(1)
(1)
(0) (0)
(1)
(1)
(0)
(1)
(1)

Fixed Point Scaling
Scale by 2 with fixed point = (2,1)
• Translate the point (2,1) to the
origin
• Scale by 2
• Translate origin to point (2,1)
(1 0 2) (2 0 0) (1 0 -2) = (2 0 -2)
(0 1 1) (0 1 0) (0 1 -1) (0 1 0)
(0 0 1) (0 0 1) (0 0 1) (0 0 1)
(2 0 -2) =
(0 1
0)
(0 0
1) =
(2) (2)
(1) (1)
(2 0 -2)
(0 1 0) (1) (1)
(0 0 1) (4) (6)
(1) (1) 0 1 2 3 4 5 6 7 8 9 10
Before
After
Chande(r1p)rabh(u1J)ainCollege of Higher Studies & School of Law

More Fixed Point Scaling
Scale by 2 with fixed point = (3,1)
• Translate the point (3,1) to the origin
• Scale by 2
• Translate origin to point (3,1)
(1 0 3) (2 0 0) (1 0 -3) = (2 0-3)
(0 1 1) (0 1 0) (0 1 -1) (0 1 0)
(0 0 1) (0 0 1) (0 0 1) (0 0 1)
(2 0 -3)
(0 1
0)
(0 0
1)
=
(2 0 -3)
(0 1
0)
(0 0
1)
=
0 1 2 3 4 5 6 7 8 9 10
Before
After
(2) (1)
(1) (1)
(1) (1)

Shears
Original
Data
y
Shear
x
Shear
1 0 0 1 b 0
a 1 0 0 1 0
0 0 1 0 0 1
GRAPHICS --> x shear --> GRAPHICS

Reflections
Reflection about the y-axis Reflection about the x-axis
-1 0 0 1 0 0
0 1 0 0 -1 0
0 0 1 0 0 1

More Reflections
Reflection about the
origin
Reflection about the line
y=x
-1 0 0 0 1 0
0 -1 0 1 0 0
0 0 1 0 0 1

UNIT-III
SOLID MODELING

UNIT-III SOLID MODELING
Start with these four, three-view drawings – Parts 1, 2, 3, 4 below and project
all points around to show alignment of one object.

Part 1 - Extrude front view (.75) inches to finish. Why front view ? Because
it shows the most of the object in solid format, and all I had to do was to add
depth onto the object. The width and height already exist. With width and
height given in front view just add depth as shown(.75).
TOP
FRONT RT. SIDE
EXTRUDE FRONT VIEW .75 INCHES TO FINISH.
WHY FRONT VIEW? BECAUSE IT SHOWS THE
MOST OF OBJECT IN SOLID FORMAT, AND ALL I
HAD TO DO WAS TO ADD DEPTH ONTO THE
OBJECT THAT ALREADY HAD WIDTH & HEIGHT.
0.75
0.75
TOP
FRONT RT. SIDE
WIDTH
DEPTH
TO FINISH OBJECT PLACE DEPTH ON
TO THE WIDTH AND HEIGHT ALREADY
GIVEN IN FRONT VIEW.

Part 2 - From the top choose this part and with a polyline extrude it (1.5)
inches by placing the height onto the depth. Move it, align it to midpoint, and
union it to form one part.
FRONT RT. SIDE
1
TOP2
From the top choose this part and with a
polyline line extrude it 1.5 inches by
placing the height on depth
Move it, align it to
midpoint and union
it to form one part.
You can do this many ways

Part 3 - What view would you pick to turn this drawing into parametricsolid?
(front view)
RT. SIDE
TOP
Part 3 - What view would you pick to make a parametric solid? .
You only need one move. Hint - place the height onto the
width.
1.00
2.50
1.00
FRO
1.00
NT
1.00

Part 4 - What part best describes this object? (1 of right side view) The command you
use is extrude. What second best part would you pick? (2 of right side view) Then also
extrude. Now move number 1 to midpoint of number 2 midpoint and union.
FRONT
RT. SIDE
PART 4 What 2 Parts would best describe the object?
The command you would use is extrude. What second
part would you pick? Then also extrude. Now move to
midpoint and union it.
2.00
1
1.50
2
If you picked 1 and 2 you are right.
0.50
TOP 1.00

Parts 1- 4 extruded & colored
UNT in partnership with TEA. Copyright ©. All rights reserved.

Drawing 2
See if you can turn these three-view drawings into solids.

Drawing 2
Lets start with top view (shows most overall shape). Extrude up (.48) inch and
place in two (.25) inch holes and subtract out the space. Place to side. Extrude
back part (2.13) inches. Move to side. Extrude part 3 back (1.50) of a inch, and
move to side. Rotate part 2 and part 3 parallel to part 1 position and stack them
on top to finish object, as shown in part 5.

Drawing 2 as Solid Model

Drawing 3 – three-view
Copy all drawings and insert into CAD,
then shade these to get better picture.

Drawing 3
Three parts - red trace in a polyline for the overall shape and extrude down (.50)
inches. Second, place in a (.25) hole and subtract out hole. Yellow draw in with a
polyline for its overall shape and extrude it (.50) of a inch. Part magenta polyline in
over shape and extrude back (.75). Place in red parallel top of yellow (as shown)
place in magenta to the midpoint of yellow (as shown) and move to right side.

Drawing 4

UNIT-IV
HIDDEN SURFACE
REMOVAL

UNIT-IV HIDDEN SURFACE REMOVAL(Visibility)
• Assumption: All polygons are opaque
• What polygons are visible with respect to your viewfrustum?
 Outside: View Frustum Clipping
Remove polygons outside of the view volume
For example, Liang-Barsky 3D Clipping
 Inside: Hidden Surface Removal
Backface culling
Polygons facing away from the viewer
Occlusion
Polygons farther away are obscured by closer polygons
Full or partially occluded portions
• Why should we remove these polygons?
Avoid unnecessary expensive operations on these polygonslater

Occlusion: Full, Partial, None
Full Partial None
• The rectangle is closer than the triangle
• Should appear in front of the triangle

Backface Culling
• Avoid drawing polygons facing away from the viewer
 Front-facing polygons occlude these polygons in a closedpolyhedron
• Test if a polygon is front- or back-facing?
front-facing
back-facing
Ideas?

Detecting Back-face Polygons
• The polygon normal of a …
front-facing polygon points towards the viewer
back-facing polygon points away from the viewer
If (n  v) > 0  “back-face”
If (n  v) ≤ 0  “front-face”
v = view vector
• Eye-space test … EASY!
“back-face” if nz <0
• glCullFace(GL_BACK)
back
front

Polygon Normals
• Let polygon vertices v0, v1, v2,..., vn - 1 be in counterclockwise
order and co-planar
• Calculate normal with cross product:
n = (v1 - v0) X (vn - 1 - v0)
• Normalize to unit vector with n/║n║
v0
v1
v2
v4
n
v3

Normal Direction
• Vertices counterclockwise  Front-facing
• Vertices clockwise  Back-facing
0
1
2
0
2
1
Front facing Back facing

Painter’s Algorithm(1)
• Assumption: Later projected polygons overwrite earlier projectedpolygons
Graphics Pipeline
1 12 23 3
Oops! The red polygon
Should be obscured by
the blue polygon

• Main Idea
 A painter creates a pictureby
drawing background scene
elemens before foreground
ones
• Requirements
 Draw polygons in back-to-
front order
 Need to sort the polygons by
depth order to get a correct
image
from Shirley

• Sort by the depth of each polygon
Graphics Pipeline
1 12 23 3
depth

• Compute zmin ranges for eachpolygon
• Project polygons with furthest zmin first
(z) depth
zmin
zmin
zmin
zmin

• Problem: Can you get a total sorting?
zmin
zmin
zmin
zmin
Correct?

• Cyclic Overlap
 How do we sort these three polygons?
• Sorting is nontrivial
 Split polygons in order to get a totalordering
 Not easy to do in general

Visibility
• How do we ensure that closer polygons overwrite further ones ingeneral?

Z-Buffer
• Depth buffer (Z-Buffer)
A secondary image buffer that holds depth values
Same pixel resolution as the color buffer
Why is it called a Z-Buffer?
After eye space, depth is simply the z-coordinate
• Sorting is done at the pixel level
Rule: Only draw a polygon at a pixel if it is closer than a
polygon that has already been drawn to this pixel

Z-BufferAlgorithm
• Visibility testing is done during rasterization

Z-buffer: A SecondaryBuffer
DAM Entertainment
Color buffer Depth buffer

Z-Buffer
P1
P2
• How do we calculate the depth values on the polygon
interior?
P3
P4
ys za zp zb
Scanline order
p
b
a
(xa  xb)
(xa  xp )
z  za  (zb  za)
(y1  y2 )
(y1  ys )
z  z1 (z2  z1)
(y1  y4 )
(y1  ys )
z  z1 (z4  z1)
Bilinear Interpolation

Z-buffer - Example
       
       
       
       
       
       
       
       
Z-buffer
Screen


 
       
       
       
       
       
       
       
       
      
     
    
   
  
 

Parallel with
the image plane





 
  
   
    
     
       
       
       
       
       
       
       
       
Not Parallel

Z-BufferAlgorithm
 Algorithm easily handles this case

Z-buffering in OpenGL
• Create depth buffer by setting GLUT_DEPTH flag in
glutInitDisplayMode()or the appropriate flag in the
PIXELFORMATDESCRIPTOR.
• Enable per-pixel depth testing with glEnable(GL_DEPTH_TEST)
• Clear depth buffer by setting GL_DEPTH_BUFFER_BIT in glClear()

Computer Graphics BCA 303

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